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\markboth{\hfil A system of differential inclusions  \hfil EJDE--2000/43}
{EJDE--2000/43\hfil Chunpeng Wang  \& Jingxue  Yin \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.~{\bf 2000}(2000), No.~43, pp.~1--8. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
%
 Uniqueness of solutions to a system of differential inclusions
\thanks{ {\em Mathematics Subject Classifications:} 35K50, 35K65, 35A05,
35D99.
\hfil\break\indent
{\em Key words:} differential inclusions, degeneracy, uniqueness.
\hfil\break\indent
\copyright 2000 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Submitted May 15, 2000. Published June 12, 2000.} }
\date{}
%
\author{ Chunpeng Wang  \& Jingxue  Yin }
\maketitle

\begin{abstract}
 In this paper we study the uniqueness of solutions to
 the initial and Dirichlet boundary-value problem of
 differential inclusions
 $$
 \Delta u_i+\nabla\cdot\stackrel{\rightarrow}{B_i}
 (u_1,u_2,\dots,u_N)\in \frac{\partial F_i(u_i)}{\partial t},
 \quad i=1,2,\dots,N,
 $$
 where $\stackrel{\rightarrow}{B_i}(s_1,s_2,\dots,s_N)$ is an
 $n$-dimensional vector continuously differentiable on ${\mathbb R}^N$,
 and $F_i(u_i)=\{w_i:u_i=A_i(w_i)\}$, $i=1,2,\dots,N$
 with $A_i(s)$ continuously differentiable functions on ${\mathbb R}$ and
 $A'_i(s)\geq 0$.
 \end{abstract}

\newcommand\iint{\int\hskip-2mm\int}
\newtheorem{theorem}{Theorem}


\section{Introduction}\label{sec1}

This paper concerns with the system of differential inclusions
$$
\Delta u_i+\nabla\cdot\stackrel{\rightarrow}{B_i}
(u_1,u_2,\dots,u_N)\in \frac{\partial F_i(u_i)}{\partial t},
\quad (x,t)\in Q_T, \quad i=1,2,\dots,N, \eqno(1.1)
$$
where $\Omega$ is a bounded domain in ${\mathbb R}^n$ with smooth boundary
$\partial\Omega$, $Q_T=\Omega\times(0,T)$, with $T>0$, $n$ and $N$ are
positive integers, $\stackrel{\rightarrow}{B_i}(s_1,s_2,\dots,s_N)$ is
an $n$-dimensional vector continuously differentiable on ${\mathbb R}^N$,
and
$$
F_i(u_i)=\{w_i:u_i=A_i(w_i)\}, \quad i=1,2,\dots,N
$$
with $A_i(s)$ continuously differentiable functions on ${\mathbb R}$ and
$A'_i(s)\geq 0$.
Note that if $A_i(s)$ is strictly increasing, then
$F_i(u_i)$ is single-valued, and (1.1) becomes equality.
However, we are interested in the case when some or all
$A_i(s)$'s are only nondecreasing, and so the $F_i(u_i)$'s are interval-valued
functions.

System (1.1) arises from mathematical models describing the
nonlinear diffusion phenomena which exist in nature extensively.
An important classical case of (1.1) is that with
$\stackrel{\rightarrow}{B_i}=\stackrel{\rightarrow}{0}$
and $N=1$. In this case (1.1) can be changed to
$$
\frac{\partial w}{\partial t}=\Delta A(w).
$$
Br\'{e}zis and Crandall \cite{B1}
 proved the uniqueness of bounded
measurable solutions for the Cauchy problem of the equation,
where the nonlinear function $A(s)$ is assumed to be only non-decreasing.
In other words, if $A(s)$ is differentiable, then
$$
A'(s)\geq 0\,;
$$
namely, the equation is permitted to be strongly degenerate.
Thereafter some authors tried to extend the uniqueness results
to the equation with convection, i.e.,
$$
\frac{\partial w}{\partial t}=\Delta A(w)
+\nabla\cdot\stackrel{\rightarrow}{B}(w).
$$
However, in most of those works, the nonlinear function $A(s)$
is assumed to be strictly increasing.
In other words, the equation is weakly degenerate,
see for example \cite{D1,C1,W2,Y1}.

In this paper we study the uniqueness of solutions of
the initial and Dirichlet boundary-value problem of (1.1).
The initial-boundary conditions are
$$
\displaylines{
\hfill u_i=0, \quad (x,t)\in \partial\Omega\times[0,T],
\quad i=1,2,\dots,N, \hfill \llap{(1.2)} \cr
\hfill F_i(u_i)(x,0)=\{f_i(x)\},\quad
x\in \Omega, \quad i=1,2,\dots,N. \hfill\llap{(1.3)}
}$$

\paragraph{Definition}
For $i=1,2,\dots,N$, let $f_i$'s be bounded and measurable functions.
$(u_1,u_2,\dots,u_N)$ is called a solution of
the initial and Dirichlet  boundary-value problem (1.1)--(1.3), if
the $u_i$'s are bounded and measurable functions and
there exist bounded measurable functions
$w_i\in F_i(u_i)$ such that for arbitrary test function
 $\varphi$ in $C^\infty({\overline{Q_T}})$ with value zero for
$x\in\partial\Omega$ and for $t=T$,
the following integral equalities hold
\begin{eqnarray*}
&&
\iint_{Q_T}
\left(u_i\Delta\varphi-\stackrel{\rightarrow}{B_i}(u_1,u_2,\dots,u_N)
\cdot\nabla\varphi+w_i\frac{\partial\varphi}{\partial t}\right)\,dx\,dt
\\
&&+\int_\Omega f_i(x)\varphi(x,0)dx=0,
\quad
i=1,2,\dots,N.
\end{eqnarray*}


The main result of this paper is the following theorem.
\begin{theorem}
The initial and Dirichlet boundary-value problem (1.1)--(1.3) has at most
one solution.
\end{theorem}

The method of the proof is inspired by Br\'{e}zis and Crandall \cite{B1}.
Here what we consider is not the Cauchy problem but
the initial and Dirichlet boundary-value problem,
so we adopt the self-adjoint
operators with homogeneous Dirichlet boundary
condition to prove the uniqueness instead of the self-adjoint operators
on the whole space.
Moreover, the problem which we consider is a system of
differential inclusions with convection,
so we must overcome some other technical difficulties.

\section{Proof of the main theorem} \label{sec2}
We first introduce a family of operators.
The $L^2$ theory for elliptic equations (see, e.g., \cite{G1}
) implies
that for each $\lambda>0$ and $f\in H^{-1}(\Omega)$, the Dirichlet problem
$$\displaylines{
\hfill -\Delta u+\lambda u=f,\quad  x\in\Omega,\hfill\llap{(2.1)} \cr
\hfill u=0,\quad x\in\partial\Omega,\hfill \llap{(2.2)}
}$$
has a unique solution $u\in H^1_0(\Omega)$.
For $0<\lambda<1$, we define the operator
$$
T_\lambda:H^{-1}(\Omega)\rightarrow H^1_0(\Omega),
\quad
f\mapsto u,
$$
where $u$ is the unique solution to (2.1)--(2.2).
It is easy to see that $T_\lambda$ is self-adjoint,
namely, for arbitrary $f,g\in H^{-1}(\Omega)$,
$$
\langle f,T_\lambda g\rangle=\langle g,T_\lambda f\rangle
$$
holds, where $\langle \cdot,\cdot\rangle$ represents
the dual product between $H^{-1}(\Omega)$ and $H^1_0(\Omega)$.
Specially, for $f\in L^2(\Omega)$ and $g\in H^{-1}(\Omega)$, we have
$$
\langle f,T_\lambda g\rangle=\int_\Omega f T_\lambda g dx.
$$
In addition, for arbitrary $f\in L^2(\Omega)$,
the $L^2$ theory for elliptic equations also implies
$T_\lambda f\in H^2(\Omega)\cap H^1_0(\Omega)$
and
$$
\|T_\lambda f\|_{H^2(\Omega)}\leq C_0 \|f\|_{L^2(\Omega)},
\eqno (2.3)
$$
here $C_0$ is a constant depending only on $n$ and $\Omega$,
but independent of $\lambda$.

\paragraph{Proof of Theorem 1.}
Let $(u_1,u_2,\dots,u_N)$ and $(\hat{u}_1,\hat{u}_2,\dots,\hat{u}_N)$
be two solutions to (1.1)--(1.3).
For $i=1,2,\dots,N$,
the bounded measurable functions in $F_i(u_i)$ and $F_i(\hat{u}_i)$
satisfying the definition are denoted by $w_i$ and
$\hat{w}_i$ correspondingly.
For $i=1,2,\dots,N$, we set
$$ \displaylines{
v_i=u_i-\hat{u}_i, \quad z_i=w_i-\hat{w}_i, \cr
\stackrel{\rightarrow}{H_i}=
\stackrel{\rightarrow}{B_i}(u_1,u_2,\dots,u_N)
-\stackrel{\rightarrow}{B_i}(\hat{u}_1,\hat{u}_2,\dots,\hat{u}_N).
}$$
The definition of solutions implies that
$z_i$, $v_i$ and $\stackrel{\rightarrow}{H_i}~(i=1,2,\dots,N)$
are all bounded measurable functions,
and for arbitrary test function $\varphi$,
namely, $\varphi\in C^\infty({\overline{Q_T}})$ with $\varphi=0$
for $x\in\partial\Omega$ and for $t=T$, the integral equalities
$$
\iint_{Q_T}
\left(v_i\Delta\varphi-\stackrel{\rightarrow}{H_i}
\cdot\nabla\varphi+z_i\frac{\partial\varphi}{\partial t}\right)\,dx\,dt=0,
\quad i=1,2,\dots,N
\eqno (2.4)
$$
hold.
Let $\psi\in C^\infty([0,T])$ with $\psi(T)=0$
and $k\in C_0^\infty(\Omega)$.
Then we see that $T_\lambda k\in H^2(\Omega)\cap H^1_0(\Omega)$.
By an approximate process, we may choose
$\psi T_\lambda k$ as a test function.
Letting $\varphi=\psi T_\lambda k$ in (2.4), we get
$$
\iint_{Q_T}
\left(\lambda\psi v_i T_\lambda k
-\psi v_i k-\psi\stackrel{\rightarrow}{H_i}
\cdot\nabla T_\lambda k
+\frac{\partial\psi}{\partial t} z_i T_\lambda k\right)\,dx\,dt=0.
$$
Using integration by parts and the self-adjointness of
$T_\lambda$, we get
$$
\iint_{Q_T}
\left(\lambda\psi k T_\lambda v_i
-\psi k v_i+\psi k T_\lambda(\nabla\cdot\stackrel{\rightarrow}{H_i})
-\psi k\frac{\partial T_\lambda z_i}{\partial t}\right)\,dx\,dt=0.
$$
Owing to the arbitrariness of  $\psi$ and $k$, we see that
$$
\frac{\partial T_\lambda z_i}{\partial t}=
\lambda T_\lambda
v_i-v_i+T_\lambda(\nabla\cdot\stackrel{\rightarrow}{H_i})
\eqno (2.5)
$$
in the sense of distribution.
It follows that $\frac{\partial T_\lambda z_i}{\partial t}\in L^2(Q_T)$
and $T_\lambda z_i\in H^2(\Omega)\cap H^1_0(\Omega)$.
Let $\psi\in C^\infty([0,T])$ with $\psi(T)=0$.
By an approximate process, we may choose $\psi T_\lambda z_i$
as a test function.
Letting $\varphi=\psi T_\lambda z_i$ in (2.4), we get
$$
\iint_{Q_T}
\left(\lambda\psi v_i T_\lambda z_i-\psi v_i z_i
-\psi\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i
+\frac{\partial\psi}{\partial t} z_i T_\lambda z_i
+\psi z_i \frac{\partial T_\lambda z_i}{\partial t}\right)\,dx\,dt=0.
\eqno (2.6)
$$
Combining (2.5) with (2.6),
we see that
\begin{eqnarray*}
\iint_{Q_T}
\Big(\lambda\psi v_i T_\lambda z_i
+\lambda\psi z_i T_\lambda v_i-2\psi v_i z_i
+\psi z_i T_\lambda(\nabla\cdot\stackrel{\rightarrow}{H_i})&& \\
-\psi\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i
+\frac{\partial\psi}{\partial t} z_i T_\lambda z_i\Big)\,dx\,dt&=&0\,.
\end{eqnarray*}
Using integration by parts and the self-adjointness of
$T_\lambda$, for $i=1,2,\dots, N$, we get
$$
\iint_{Q_T}
\Big(2\lambda\psi v_i T_\lambda z_i-2\psi v_i z_i
-2\psi\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i
+\frac{\partial\psi}{\partial t} z_i T_\lambda z_i\Big)\,dx\,dt=0.
\eqno (2.7)
$$
Let
$$
g_{i\lambda}(t)=\int_\Omega z_i T_\lambda z_i dx,
\quad t\in[0,T],
\quad i=1,2,\dots,N.
$$
Now we prove that $g_{i\lambda}(t)$ converges to zero
on $[0,T]$ uniformly as $\lambda\rightarrow 0$ for $i=1,2,\dots,N$.

First, we show that $g_{i\lambda}(t)$ is absolutely continuous.
From (2.7), we get
$$
\iint_{Q_T}\left(2\lambda\psi v_i T_\lambda z_i-2\psi v_i z_i
-2\psi\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i
-\psi\frac{\partial (z_i T_\lambda z_i)}{\partial t}\right)\,dx\,dt=0.
$$
From the arbitrariness of  $\psi$, we see that
\begin{eqnarray*}
g'_{i\lambda}(t)
&=&\frac{d}{dt}\int_\Omega z_i T_\lambda z_i dx
=\int_\Omega\frac{\partial (z_i T_\lambda z_i)}{\partial t} \,dx\\
&=&2\lambda\int_\Omega v_i T_\lambda z_i dx
-2\int_\Omega v_i z_i dx
-2\int_\Omega\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i dx,
\quad\hbox{a.e.}~t\in [0,T].
\end{eqnarray*}
Since $z_i$, $v_i$ and $\stackrel{\rightarrow}{H_i}$ are all bounded
measurable functions and (2.3) holds,
we get that $g'_{i\lambda}(t)\in L^1(0,T)$.
Thus $g_{i\lambda}(t)$ is absolutely continuous.

Next, we show that $g_{i\lambda}(0+0)
\equiv\lim_{t\rightarrow 0^+}g_{i\lambda}(t)=0$.
Let
$$
\psi_\varepsilon(t)=\int^{+\infty}_t
\alpha_\varepsilon(s-\varepsilon)ds,
\quad \alpha_\varepsilon(s)=\frac1\varepsilon
\alpha\left(\frac{s}{\varepsilon}\right),
$$
where $\alpha(s)$ denotes the kernel of one-dimensional mollifier, namely,
$\alpha$ is in the space $C_0^\infty(-\infty,+\infty)$, $\alpha\geq0$,
supp$\alpha=[-1,1]$ and $\int_{-1}^1\alpha(s)\,ds=1$.
Thus $\psi_\varepsilon\in C^\infty([0,T])$ and $\psi_\varepsilon(T)=0$
for sufficiently small $\varepsilon>0$.
Letting $\psi=\psi_\varepsilon$ in (2.7), we get
$$
\iint_{Q_T}
(2\lambda\psi_\varepsilon v_i T_\lambda z_i
-2\psi_\varepsilon v_i z_i
-2\psi_\varepsilon\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i
-\alpha_\varepsilon(t-\varepsilon) z_i T_\lambda z_i)\,dx\,dt=0.
$$
The dominated convergence theorem implies
\begin{eqnarray*}
g_{i\lambda}(0+0)
&=& \lim_{\varepsilon\rightarrow0^+}\int^{2\varepsilon}_0
\alpha_\varepsilon(t-\varepsilon)g_{i\lambda}(t)dt \\
&=& \lim_{\varepsilon\rightarrow0^+}\iint_{Q_T}
\alpha_\varepsilon(t-\varepsilon) z_i T_\lambda z_i \,dx\,dt\\
&=& 2\lambda\lim_{\varepsilon\rightarrow 0^+}\iint_{Q_T}
\psi_\varepsilon v_i T_\lambda z_i \,dx\,dt
-2\lim_{\varepsilon\rightarrow 0^+}\iint_{Q_T}
\psi_\varepsilon v_i z_i \,dx\,dt\\
&&-2\lim_{\varepsilon\rightarrow 0^+}\iint_{Q_T}
\psi_\varepsilon\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i \,dx\,dt\,.
\end{eqnarray*}
Since $z_i$, $v_i$ and $\stackrel{\rightarrow}{H_i}$ are all bounded
measurable functions and (2.3) holds,
we get that $g_{i\lambda}(0+0)=0$.

Finally, we prove that $g_{i\lambda}(t)$ converges to zero
on $[0,T]$ uniformly as $\lambda\rightarrow 0$.
It follows easily from the above arguments that
\begin{eqnarray*}
g_{i\lambda}(t)
&=&g_{i\lambda}(0+0)+\int^t_0 g'_{i\lambda}(s)ds\\
&=& 2\lambda\int_0^t\int_\Omega v_i T_\lambda z_i \,dx\,ds
-2\int_0^t\int_\Omega v_i z_i \,dx\,ds\\
&& -2\int_0^t\int_\Omega\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i
\,dx\,ds\,.
\end{eqnarray*}
Since $w_i$ and $\hat{w}_i$ are bounded measurable and
$A_i$ and $\stackrel{\rightarrow}{B_i}$ are continuously differentiable,
there exist three positive constants $M_0$, $M_1$ and $M_2$
such that for $i=1,2,\dots,N$, the following estimates hold
$$
|z_i|\leq M_0, \quad |v_i|\leq M_1|z_i|,
\quad |\stackrel{\rightarrow}{H_i}|
\leq M_2\left(\sum_{j=1}^N v_j^2\right)^{1/2}.
$$
Noticing that $z_i$ and $v_i$ have the same sign for $A'_i(s)\geq0$,
we get
$$
v_i z_i=|v_i||z_i|\geq\frac{1}{M_1}v_i^2.
$$
By Schwarz's inequality and Young's inequality, we get
\begin{eqnarray*}
\lefteqn{ \big|\int_\Omega\stackrel{\rightarrow}{H_i}\cdot\nabla T_\lambda z_i
\,dx \big| } \\
&\leq&\left(\int_\Omega|\stackrel{\rightarrow}{H_i}|^2 dx\right)^{1/2}
\left(\int_\Omega|\nabla T_\lambda z_i|^2 dx\right)^{1/2}\\
&\leq& M_2\sum_{j=1}^N\left(\int_\Omega v_j^2 dx\right)^{1/2}
\left(\int_\Omega|\nabla T_\lambda z_i|^2 dx\right)^{1/2}\\
&\leq&
\frac{1}{N M_1}\sum_{j=1}^N\int_\Omega v_j^2 dx+
\frac{N^2 M_1 M_2^2}{4}
\int_\Omega(\nabla T_\lambda z_i\nabla T_\lambda z_i)dx\\
&=&
\frac{1}{N M_1}\sum_{j=1}^N\int_\Omega v_j^2 dx+
\frac{N^2 M_1 M_2^2}{4}
\int_\Omega(-T_\lambda z_i\Delta T_\lambda z_i)dx\\
&=&
\frac{1}{N M_1}\sum_{j=1}^N\int_\Omega v_j^2 dx+
\frac{N^2 M_1 M_2^2}{4}\int_\Omega\left(-\lambda (T_\lambda z_i)^2
+z_i T_\lambda z_i\right) dx\\
&\leq&
\frac{1}{N M_1}\sum_{j=1}^N\int_\Omega v_j^2 dx+
\frac{N^2 M_1 M_2^2}{4}g_{i\lambda}(t).
\end{eqnarray*}
Let
$$
g_\lambda(t)=\sum_{i=1}^N g_{i\lambda}(t).
$$
Therefore,
\begin{eqnarray*}
\lefteqn{g_\lambda(t)}\\
&=& 2\sum_{i=1}^N\Big(\lambda\int_0^t\int_\Omega v_i T_\lambda z_i \,dx\,ds
-\int_0^t\int_\Omega v_i z_i \,dx\,ds
-\int_0^t\int_\Omega\stackrel{\rightarrow}{H_i}
\cdot\nabla T_\lambda z_i \,dx\,ds\Big)\\
&\leq&
2\sum_{i=1}^N\Big(\lambda\int_0^t\int_\Omega v_i T_\lambda z_i \,dx\,ds
-\frac{1}{M_1}\int_0^t\int_\Omega v_i^2 \,dx\,ds\\
&&
+\frac{1}{N M_1}\sum_{j=1}^N\int_0^t\int_\Omega v_j^2 \,dx\,ds
+\frac{N^2 M_1 M_2^2}{4}\int_0^tg_{i\lambda}(s)ds\Big)\\
&\leq&
2\lambda\sum_{i=1}^N\int_0^T\int_\Omega|v_i T_\lambda z_i|\,dx\,ds
+\frac{N^2 M_1 M_2^2}{2}\int_0^t g_\lambda(s)ds.
\end{eqnarray*}
Moreover, it follows that
$$
g_\lambda(t)\geq0
$$
by
\begin{eqnarray*}
g_{i\lambda}(t)
&=&\int_\Omega z_i T_\lambda z_i dx\\
 &=& \int_\Omega(-\Delta T_\lambda z_i T_\lambda z_i+
\lambda T_\lambda z_i T_\lambda z_i)dx\\
&=&
\int_\Omega(\nabla T_\lambda z_i \nabla T_\lambda z_i+
\lambda T_\lambda z_i T_\lambda z_i)dx\\
&\geq&0\,.
\end{eqnarray*}
Hence by Gronwall's inequality, we get
$$
g_\lambda(t)\leq C_1\lambda,
$$
where $C_1$ is a constant depending only on $N$, $M_0$, $M_1$,
$M_2$, $C_0$, $T$ and the measure of $\Omega$,
but independent of $\lambda$ and $t$.
So $g_\lambda(t)$ converges to zero
on $[0,T]$ uniformly as $\lambda\rightarrow 0$.
Noticing that $g_{i\lambda}(t)\geq0$, we get that $g_{i\lambda}(t)$
converges to zero on $[0,T]$ uniformly as $\lambda\rightarrow 0$.

Now we prove
$$
z_i(x,t)=0,
\quad
\hbox{a.e.}~(x,t)\in Q_T,
\quad
i=1,2,\dots,N.
$$
For any $\varphi\in C^\infty_0(Q_T)$, we have
\begin{eqnarray*}
\lefteqn{\big|\iint_{Q_T}z_i\varphi \,dx\,dt\big|^2}\\
&=&
\left|\iint_{Q_T}
(-\Delta T_\lambda z_i +\lambda  T_\lambda z_i) \varphi \,dx\,dt\right|^2\\
&=&
\left|\iint_{Q_T}(\nabla T_\lambda z_i \nabla\varphi
+\lambda\varphi T_\lambda z_i) \,dx\,dt\right|^2\\
&\leq&
2\|\nabla\varphi\|_{L^2(Q_T)}^2\|\nabla T_\lambda z_i\|_{L^2(Q_T)}^2
+2\lambda^2\|\varphi\|_{L^2(Q_T)}^2\|T_\lambda z_i\|_{L^2(Q_T)}^2\\
&\leq&
C_2
\left(\iint_{Q_T}|\nabla T_\lambda z_i|^2 \,dx\,dt+
\lambda\iint_{Q_T}(T_\lambda z_i)^2 \,dx\,dt\right)\\
&\leq&
C_2\iint_{Q_T} (-\Delta T_\lambda z_i+
\lambda T_\lambda z_i)T_\lambda z_i \,dx\,dt\\
&=&
C_2\iint_{Q_T}z_i T_\lambda z_i\,dx\,dt\\
&\leq&
C_2 T\sup_{t\in[0,T]}g_{i\lambda}(t)\rightarrow 0,
\quad
(\lambda\rightarrow 0),
\end{eqnarray*}
where $C_2=2\|\nabla\varphi\|_{L^2(Q_T)}^2+2\|\varphi\|_{L^2(Q_T)}^2$
independent of $\lambda$.
Therefore,
$$
\iint_{Q_T}z_i\varphi \,dx\,dt=0,
\quad \forall \varphi\in C^\infty_0(Q_T).
$$
It follows that
$$
z_i(x,t)=0,
\quad \hbox{a.e.}~(x,t)\in Q_T,\quad i=1,2,\dots,N.
$$
Thus
$$
w_i(x,t)=\hat{w}_i(x,t),
\quad \hbox{a.e.}~(x,t)\in Q_T, \quad i=1,2,\dots,N,
$$
which implies
$$
u_i(x,t)=\hat{u}_i(x,t),
\quad \hbox{a.e.}~(x,t)\in Q_T,\quad i=1,2,\dots,N.
$$
The proof is complete.

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\end{thebibliography} \medskip


\noindent{\sc Chunpeng Wang  \& Jingxue  Yin} \\
Department of Mathematics, JiLin University, \\
Changchun, Jilin 130023, People's Republic of China\\
e-mail: yjx@mail.jlu.edu.cn

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